Factorisation

First (x^4+2x^2y^2+y^4=(x^2+y^2)^2), and then (16=4^2). Exam tip: solve mixed identities in two steps.

First write (x^4-16) as ((x^2)^2-4^2) and then factor (x^2-4). Exam tip: do not stop before the complete factorisation.

In (a^4-b^4), first use difference of squares and then factor (a^2-b^2). Exam tip: (a^2+b^2) is usually not factored further here.

The first three terms form a perfect square because they match the identity \(u^2+2uv+v^2=(u+v)^2\). Taking \(u=x\) and \(v=y\), we get \(x^2+2xy+y^2=(x+y)^2\). The remaining term is \(9z^2=(3z)^2\), so the whole expression becomes the difference of the squares \((x+y)^2-(3z)^2\). The correct method is therefore to identify the perfect square first and then use the difference-of-squares identity.

Using \(A^2-B^2=(A-B)(A+B)\), we obtain \((x+y-3z)(x+y+3z)\), which is option B. Expanding confirms it: the product gives \((x+y)^2-9z^2\), and then \((x+y)^2=x^2+2xy+y^2\). Option A uses \(x-y\), which would change the sign of the middle term. Options C and D do not produce the required terms, so they cannot be correct.

The direct answer is option D: (2p−3q−5r)(2p−3q+5r). First group the expression as (4p²−12pq+9q²)−25r². The first group is (2p−3q)² because its terms are (2p)²−2(2p)(3q)+(3q)². Also, 25r²=(5r)². Thus the whole expression is A²−B² with A=2p−3q and B=5r, so it factors as (A−B)(A+B), giving option D. Option A would use (2p+3q)² and would produce a positive 12pq, so its sign is wrong. Option B leaves the difference of squares partly unfactored, so it is not the complete factorised form requested. Option C does not expand to the original expression. D is exact. Memory cue: first complete the three-term square, then apply difference of squares.

Write (x^3-8y^3) as ((x-2y)(x^2+2xy+4y^2)) and take the common factor. Exam tip: identify the common binomial first.

The direct answer is option B: \((2x+y)(x+3y)\). To factor, multiply the first terms to get \(2x^2\) and the last terms to get \(3y^2\). The middle products are \(6xy\) and \(xy\), whose sum is \(7xy\). Thus \(2x^2+6xy+xy+3y^2=2x^2+7xy+3y^2\). Option A gives \(2x^2+5xy+3y^2\), so its middle term is wrong. Option B gives exactly the expression, so it is correct. Option C gives \(2x^2+7xy+6y^2\), so its last term is wrong. Option D gives \(2x^2-7xy+3y^2\), so the middle sign is wrong. Memory cue: for \(ac=6\) and sum \(7\), use \(1\) and \(6\).

Expanding gives ( (3x+4)(2x-3)=6x^2-x-12 ). Exam tip: verify by expansion when signs are tricky.

For (-7ab), write (12a^2-8ab+ab-b^2). Exam tip: when options look similar, confirm by expansion.

Treat (x^2) as a new term and factor like a quadratic. Exam tip: for (x^4) type questions, think (u=x^2).

Taking (y^2) as one quantity, the product must be (36) and the sum (-13). Exam tip: identify the pair (-4) and (-9).

All middle terms form (2ab)-type pairs of (a), (2b), and (3c). Exam tip: check a three-term square pairwise.

This is the perfect square of (m), (-n), and (p). Exam tip: decide bracket signs from the signs of mixed terms.

The numbers (9) and (-5) have sum (4) and product (-45). Exam tip: with opposite signs, the larger magnitude gives the middle term sign.

Expanding gives ( (3x-2)(x-5)=3x^2-17x+10 ). Exam tip: choose the pair with product (30) and sum (-17).

In \(ax+ay+bx+by\), group the terms as \(a(x+y)+b(x+y)\), then take \((x+y)\) common to get \((a+b)(x+y)\). \(x^2-16\) instead uses the difference-of-squares identity. Exam tip: look for a repeated bracket after grouping.

Expanding \((7a+2b)(a-b)\) gives \(7a^2-7ab+2ab-2b^2=7a^2-5ab-2b^2\). Hence, option A is correct. Option B produces a middle term of \(+5ab\), not \(-5ab\). Exam tip: expand the factors and check the sign of the middle term carefully.

This is the identity (x^3+y^3+z^3-3xyz=(x+y+z)(x^2+y^2+z^2-xy-yz-zx)). Exam tip: remember the three-variable cubic identity.

In \(a^2+6ab+9b^2\), the first and last terms are \(a^2\) and \((3b)^2\). The middle term is \(2\times a\times3b=6ab\), so it equals \((a+3b)^2\). Exam tip: always verify the middle coefficient.

It is ( (4x)^3+(5y)^3+(6z)^3-3(4x)(5y)(6z) ). Exam tip: apply the identity (a^3+b^3+c^3-3abc).

In \(49p^2-q^2\), \(49p^2=(7p)^2\) and \(q^2\) is also a perfect square, with a minus sign between them. Hence it is a difference of squares. Options A and D are perfect-square trinomials. Exam tip: check for two square terms separated by ‘−’.

First take out common factor (2) and apply sum of cubes to (x^3+8). Exam tip: do not forget the common factor.

\(9x^2-16y^2=(3x)^2-(4y)^2\), so it fits \(a^2-b^2=(a-b)(a+b)\): \((3x-4y)(3x+4y)\). Option B has a sum, not a difference. Exam tip: check for two perfect squares separated by a minus sign.

First we get ((x^2-1)(x^2-9)), and both are differences of squares. Exam tip: factor till the final complete form.

The direct answer is D: \((3x+5y-7z)(3x+5y+7z)\). First group the first three terms: \(9x^2+30xy+25y^2=(3x+5y)^2\), because the middle term is \(2(3x)(5y)=30xy\). Also, \(49z^2=(7z)^2\). Thus the expression is \((3x+5y)^2-(7z)^2\), which factors as the difference of squares into option D. Option A uses \(3x-5y\), producing \(-30xy\), not plus. Option B is not the needed square-difference structure. Option C is a square and cannot produce the subtraction of \(49z^2\). Option D expands correctly. First make a perfect square, then use difference of squares.